Question

A parallel-plate capacitor, with plate area $A$ and separation between the plates $d$, is charged by a constant current $i$. Consider a plane surface of area $\frac{A}{2}$ parallel to the plates and drawn symmetrically between the plates. Find the displacement current through this area.

Answer 1

Suppose the charge on the capacitor at time $t$ is $Q$.

The electric field between the plates of the capacitor is,

$E=\frac{Q}{{\epsilon }_{0}A}$

The flux through the area considered is,

${\varphi }_E=E\times \frac{A}{2}=\frac{Q}{{\epsilon }_{0}A}\times \frac{A}{2}=\frac{Q}{2{\epsilon }_0}$

The displacement current is given by

$I_d={\epsilon }_0\frac{d{\varphi }_E}{dt}$

$I_d={\epsilon }_0\left(\frac{1}{2{\epsilon }_0}\right)\frac{dQ}{dt}$

$\Rightarrow I_d=\frac{i}{2}$

Hence, option $\left(B\right)$ is correct.